CONSTRUCTIONS PRESERVING HILBERT SPACE UNIFORM EMBEDDABILITY OF DISCRETE GROUPS

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1 CONSTRUCTIONS PRESERVING HILBERT SPACE UNIFORM EMBEDDABILITY OF DISCRETE GROUPS MARIUS DADARLAT AND ERIK GUENTNER Abstract. Uniform embeddability (in a Hilbert space), introduced by Gromov, is a geometric property of metric spaces. As applied to countable discrete groups, it has important consequences to the Novikov conjecture. Exactness, introduced and studied extensively by Kirchberg-Wassermann, is a functional analytic property of locally compact groups. Recently it has become apparent that as properties of countable discrete groups uniform embeddability and exactness are closely related. We further develop the parallel between these classes by proving that the class of uniformly embeddable groups shares a number of permanence properties with the class of exact groups. In particular, we prove that it is closed under direct and free products (with and without amalgam), inductive limits and certain extensions. 1. Introduction Gromov introduced the notion of uniform embeddability of metric spaces and suggested that finitely generated discrete groups that are uniformly embeddable in a Hilbert space, when viewed as metric spaces, might satisfy the Novikov Conjecture [12, 10]. Yu proved that this is indeed the case [20, 18]. Kirchberg-Wassermann defined the notion of exactness of a locally compact group in terms of the behavior of its reduced crossed product functor. Subsequently, they developed the main properties of exact groups. In particular they showed that in the case of a countable discrete group, exactness can be reformulated entirely in terms of the reduced C -algebra [14], that is, that exactness is a property of the harmonic analysis of the left regular representation of such a group. The starting point of this work is the startling fact that for countable discrete groups, uniform embeddability (in a Hilbert space, a geometric property) and exactness (an analytic property) are closely related. The first indications of the relationship between uniform Date: July 18, The first author was supported in part by an MSRI Research Professorship and NSF Grant DMS The second author was supported in part by an MSRI Postdoctoral Fellowship and NSF Grant DMS

2 2 MARIUS DADARLAT AND ERIK GUENTNER embeddability and exactness are found in the work of Guentner-Kaminker [13]; these preliminary steps were quickly expanded by Ozawa [16], Anantharaman-Delaroche [2] and others. We are concerned with uniformly embeddable groups. Our main results are outlined in the following theorem (more precise statements follow in later sections) which summarizes the basic permanence properties of the class of uniformly embeddable groups. Observe that the properties described are all shared by the class of countable discrete exact groups [15]. Indeed, in each case it is possible to give a unified account of the results for uniform embeddability and exactness; in some cases our methods provide alternate proofs of the known results concerning exactness. Theorem. The class of countable discrete groups that are uniformly embeddable in a Hilbert space is closed under subgroups and products, direct limits, free products with amalgam, and extensions by exact groups. The fact that subgroups and products of uniformly embeddable groups are again uniformly embeddable is elementary and quite well-known; they are included in the statement for completeness. The other properties are more difficult to establish. It is possible to construct a uniform embedding of a free product (without amalgam) directly from uniform embeddings of the factors [5]. On the other hand, the corresponding result for free products with amalgam is considerably more difficult in view of the fact that the common subgroup of the amalgam can introduce considerable distortion into the product. Our proof is based on a suitable adaptation of an argument given by Tu in his work on Property A [19]; although we are not able to verify a number of assertions concerning the metric defined in Section 9 in Tu s paper we are able to adapt his arguments to the present context. The proof we give works equally well for countable exact groups, and is unrelated to Dykema s original proof that the class of countable exact groups is closed under free products with amalgam [9, 8]. Again, in the case without amalgam a considerably simpler proof of this fact is now available [5]. The general problem of uniform embeddability of extensions is intriguing. Our proof that the class of uniformly embeddable groups is closed under extensions by exact groups is inspired by the argument of Anantharaman-Delaroche-Renault showing that the class of countable exact groups is closed under extensions [3]. It is unknown whether the class of uniformly embeddable groups is closed under general extensions; even the case of a central extension of Z by a uniformly embeddable group remains open. At present, the behavior with respect to extensions provides the best possibility of distinguishing the classes of uniformly embeddable and exact groups.

3 UNIFORM EMBEDDABILITY 3 We draw two immediate corollaries. Since they are peripheral to our study we will not establish notation or provide the relevant definitions; rather, we provide references. Corollary. The class of countable discrete groups that are uniformly embeddable in a Hilbert space is closed under the formation of HNN extensions. Proof. An HNN extension is built from free products with amalgam, direct limits, and a semi-direct product by Z, which is exact [17, 4]. (See [6] for related results.) Corollary. The fundamental group of a graph of countable discrete groups is uniformly embeddable in a Hilbert space if and only if each of the groups is uniformly embeddable in a Hilbert space. Proof. Each constituent group is a subgroup of the fundamental group. Conversely, the fundamental group of a graph of groups is built from free products with amalgam, HNN extensions and direct limits [17, 4]. (See [1] for related remarks.) 2. Background Let X and Y be metric spaces, with metrics d X and d Y, respectively. A function F : X Y is a uniform embedding if there exist non-decreasing functions ρ ± : R + R + such that lim t ρ ± (t) = and such that ρ (d X (x, x )) d Y (F (x), F (x )) ρ + (d X (x, x )), for all x, x X. (1) The space X is uniformly embeddable if there exists a uniform embedding F of X into a Hilbert space H. Uniform embeddability in a real Hilbert space is equivalent to uniform embeddability in a complex Hilbert space; henceforth we shall deal only with real Hilbert spaces. Obviously, if X is countable we may assume that the Hilbert space is separable. A metric space X is locally finite if for every x X and R > 0 the metric ball with center x and radius R is finite. In this case the metric is called proper. A locally finite metric space is discrete (as a topological space in the metric topology). In the case of locally finite metric spaces there are a number of equivalent formulations of uniform embeddability [7, 13]; to these we add the following simple extension, which applies to any metric space and which will be our fundamental criterion for uniform embeddability. Proposition 2.1. Let X be a metric space. Then X is uniformly embeddable if and only if for every R > 0 and ε > 0 there exists a Hilbert space valued map ξ : X H, (ξ x ) x X, such that ξ x = 1, for all x X and such that

4 4 MARIUS DADARLAT AND ERIK GUENTNER (i) sup{ ξ x ξ x : d(x, x ) R, x, x X} ε (ii) lim S sup{ ξ x, ξ x : d(x, x ) S, x, x X} = 0. These conditions may be replaced by (iii) sup{ 1 ξ x, ξ x : d(x, x ) R, x, x X} ε (iv) lim S inf{ ξ x ξ x : d(x, x ) S, x, x X} = 2, respectively. Remark. We refer to (i) and (iii) collectively as the convergence condition; similarly, we refer to (ii) and (iv) collectively as the support condition. Proof. The interchangeability of (i) (iii) and (ii) (iv) follows from the simple observation that for unit vectors ξ, η H we have ξ η 2 = 2 2 ξ, η. Assume that X is uniformly embeddable and let F : X H be a uniform embedding in a real Hilbert space H. Let and define Exp : H Exp(H) by Exp(H) = R H (H H) (H H H) Exp(ζ) = 1 ζ ( ) ( ) 1 1 ζ ζ ζ ζ ζ. 2! 3! Note that Exp(ζ), Exp(ζ ) = e ζ,ζ, for all ζ, ζ H. For t > 0 define t F (x) 2 ξ x = e Exp( 2t F (x)). It is easily verified that ξ x, ξ x = e t F (x) F (x ) 2. Consequently, for all x, x X we have ξ x = 1, and e tρ +(d(x,x )) 2 ξ x, ξ x e tρ (d(x,x )) 2. (2) Letting t = ε(1 + ρ + (R) 2 ) 1 it is easy to verify the conditions (iii) and (ii) above. Conversely, assume that X satisfies the conditions in the second part of the statement. There exists a sequence of maps η n : X H n and a sequence of numbers S 0 = 0 < S 1 < S 2 <... increasing to infinity such that for every n 1 and every x, x X (i) η n (x) = 1, (ii) η n (x) η n (x ) 1/n, provided d(x, x ) n (iii) η n (x) η n (x ) 1, provided d(x, x ) S n. Choose a base point x 0 X and define F : X n=1h n by F (x) = 1 2 (η 1(x) η 1 (x 0 ) η 2 (x) η 2 (x 0 ) ).

5 It is not hard to verify that F is well defined and UNIFORM EMBEDDABILITY 5 ρ (d(x, x )) F (x) F (x ) d(x, x ) + 1, for all x, x X, where ρ = 1 2 n=1 n 1 χ[sn 1,S n), and the χ [Sn 1,S n) are the characteristic functions of the sets [S n 1, S n ). Indeed, let x, x X. If n is such that n 1 d(x, x ) < n we have F (x) F (x ) 2 = 1 η i (x) η i (x ) η i (x) η i (x ) i n 1 (n 1) i n 1 i n i n 1 i 2 d(x, x ) Similarly, if n is such that S n 1 d(x, x ) < S n we have F (x) F (x ) 2 1 η i (x) η i (x ) 2 n 1 = ρ (d(x, x )) Remark. Straightforward modifications of the above argument produce a uniform embedding F satisfying the sharper estimate F (x) F (x ) d(x, x ) + δ, for an arbitrarily chosen δ > 0; simply replace the 1/n in (ii) by δ/2 n. Two metrics d and d on the set X are coarsely equivalent if for every R > 0 there exists an S > 0 such that the d metric ball with center x and radius R is contained in the d metric ball with center x and radius S; and conversely. Equivalently, two metrics on X are coarsely equivalent if the identity map X X is a uniform embedding. Proposition 2.2. Let d and d be coarsely equivalent metrics on X. Then X is uniformly embeddable with respect to d if and only if it is uniformly embeddable with respect to d. Proof. If two maps are uniform embeddings, so is their composition. Remark. Below we require only the fact that if Y H and X Y are uniform embeddings then the composite X H is a uniform embedding. In other words, we do not need to know that X and Y are coarsely equivalent to conclude the uniform embeddability of X from that of Y. Let Γ be a countable discrete group. A length function on Γ is a non-negative, real-valued function l satisfying, for all a and b Γ, (i) l(ab) l(a) + l(b),

6 6 MARIUS DADARLAT AND ERIK GUENTNER (ii) l(a 1 ) = l(a), (iii) l(a) = 0 if and only if a = 1, A length function l is proper if for all C > 0 the subset l 1 ([0, C]) Γ is finite. One can construct an integer-valued proper length function on Γ as follows. Let S be a symmetric set of generators of Γ. Let l 0 : S N be a proper function satisfying (ii) (iii) above. Then l(a) = inf{l 0 (a 1 ) + + l 0 (a n ) : a = a 1 a n, a i S} is a proper length function on Γ. Given a length function l we define a metric d by d(a, b) = l(a 1 b). A metric constructed in this way from a length function is left invariant in the sense that d(ca, cb) = d(a, b), for all a, b and c Γ. Conversely, every left invariant metric arises in this way from a length function. A length function is proper if and only if the corresponding (left invariant) metric has bounded geometry. Recall that a metric space X has bounded geometry if for every R > 0, there is a uniform bound on the number of elements in the balls of radius R in X. We require the following well-known proposition (compare [19, Lemma 2.1]). Proposition 2.3. Let Γ be a countable discrete group and let d and d be metrics on Γ associated to proper length functions l and l, respectively. Then d and d are coarsely equivalent. Proof. Since the metrics are left invariant it suffices to consider the containment, as in the definition, of balls centered at the identity element. By symmetry it suffices to show that for every R > 0 there exists an S > 0 such that for all a Γ if l(a) < R then l (a) < S. But, since l is proper this is obvious. As a consequence of the previous two propositions, uniform embeddability of a countable discrete group Γ is independent of the particular proper length function used to define its metric. Consequently, we systematically omit reference to a specific length function or metric in the statements of all results and say simply Γ is uniformly embeddable to mean that Γ is uniformly embeddable in a Hilbert space for some (equivalently all) left invariant proper metric(s). Finally, we draw two simple consequences. An action of a discrete group on a locally finite metric space X is proper if for every bounded subset B X the set { a Γ : a B B } is finite. Equivalently, for every x X and R 0 the set { a Γ : a x B R (x) } is finite. Observe that a free action of a discrete group on a locally finite metric space is proper.

7 UNIFORM EMBEDDABILITY 7 Corollary 2.4. Let Γ be a countable discrete group equipped with a left invariant proper metric. Let X be a locally finite metric space equipped with a free isometric action of Γ. Then the inclusion Γ X as an orbit is a uniform embedding. Proof. Let x 0 X. Since the action of Γ on X is by isometries, l(a) = d X (a x 0, x 0 ) defines a length function on Γ. Since the action is free and X is locally finite, l is a proper length function. Let d be the left invariant metric associated to l. According to the previous proposition, the original metric on Γ is coarsely equivalent to d, which is precisely what was to be proved. Corollary 2.5. Let X and Γ be as in the statement of the previous corollary. uniformly embeddable then so is Γ. If X is Property A is a condition on metric spaces introduced by Yu [20]. We will work with the following characterization of Property A. Proposition 2.6 ([19]). A discrete metric space X with bounded geometry has Property A if and only if for every R > 0 and ε > 0 there exists an S > 0 and a Hilbert space valued function ξ : X H such that for all x, x X we have ξ x = 1, and (i) d(x, x ) R ξ x ξ x ε (ii) d(x, x ) S ξ x, ξ x = 0 Equivalently, for every R > 0 and ε > 0 there exists an S > 0 and ξ : X l 2 (X) such that for all x, x X we have ξ x = 1, (i) as above and (iii) supp ξ x B S (x). Remark. As in the case of uniform embeddability we refer to (i) as the convergence condition and to (ii) and (iii) collectively as the support condition. It is clear from the proposition that Property A is invariant under coarse equivalence. We refer the reader to [15] for an introduction to exact groups. Our interest is Prop. 2.6 is motivated by the following result inspired by [13]. Theorem 2.7 ([16], and also [2]). A countable discrete group Γ is exact if and only if Γ has Property A with respect to some (every) left-invariant proper metric. In analogy with Cor. 2.5 we have the following result, in which we do not assume that X itself has Property A.

8 8 MARIUS DADARLAT AND ERIK GUENTNER Corollary 2.8. Let Γ be a countable discrete group. Assume that Γ acts freely and isometrically on a locally finite metric space X satisfying the convergence and support conditions of Prop Then Γ is exact. Proof. Include Γ X as an orbit. The convergence and support conditions of Prop. 2.6 pass from X to the subspace Γ. Since Γ has bounded geometry Prop. 2.6 applies. 3. Limits We begin to establish the closure properties of the class of countable discrete uniformly embeddable groups. In this section we treat direct limits, our main result being the following proposition. Proposition 3.1. Let Γ be the limit of a directed system of countable discrete groups G 1 G 2 G 3 in which the the maps G n G n+1 are injective. If each of the groups G n is uniformly embeddable then so is Γ. Proof. The proof is based on a method of extending Hilbert space valued functions from a subgroup to an ambient group. Specifically, let ξ : G H be a Hilbert space valued function on a subgroup G of a countable discrete group Γ. Choose and fix a family of coset representatives x X Γ for Γ/G; having done so each element a Γ can be uniquely expressed as a product x a g a, where x a X and g a G. The extension ξ of ξ is defined by ξ : Γ H l 2 (Γ/G) = H l 2 (X), ξa = ξ ga δ ag = ξga δ xa. Let now Γ be a direct limit as in the statement of the theorem. Equip Γ with a proper length function l Γ and associated metric d Γ. Metrize each of the subgroups G n as subspaces of Γ; the metric and length function on G n are simply the restriction of d Γ and l Γ. Observe that for a, b Γ we have and that in this case, a 1 b G n ag n = bg n x a = x b Γ, d Gn (g a, g b ) = d Γ (g a, g b ) = d Γ (x a g a, x b g b ) = d Γ (a, b). (3) We show that Γ satisfies the convergence and support conditions of Prop Let ε > 0, and R > 0 be given. Obtain n such that if a Γ has l Γ (a) R then a G n ; this is possible because the length function l Γ on Γ is assumed to be proper. According to the criterion for uniform embeddability obtain a Hilbert space valued function ξ : G n H and such that for all g, h G n we have ξ g = 1 and

9 UNIFORM EMBEDDABILITY 9 (i) if d Gn (g, h) R then ξ g ξ h < ε, (ii) ˆε > 0 S > 0 such that if d Gn (g, h) S then ξ g, ξ h < ˆε. Let ξ be the extension of ξ to Γ defined above. Clearly, ξ a = 1 for all a Γ and it remains to verify the conditions of Prop Let a, b Γ. For the convergence condition assume that l Γ (a 1 b) = d Γ (a, b) R. By our choice of n we have a 1 b G n and, according to (3), d Gn (g a, g b ) = d Γ (a, b) R. Therefore, ξ a ξ b = ξ ga δ agn ξ gb δ bgn = ξ ga ξ gb < ε. For the support condition let ˆε > 0, obtain S as in (ii) above and assume d Γ (a, b) S. According to ξ a, ξ ξ ga, ξ gb, if ag n = bg n b = ξ ga, ξ gb δ agn, δ bgn = (4) 0, otherwise. we may assume a 1 b G n. But then, according to (3) again, d Gn (g a, g b ) = d Γ (a, b) and ξ a, ξ b < ˆε. Remark. The previous argument is easily adjusted to yield a new proof of the fact that a direct limit, as in the statement of the proposition, of exact groups is again exact [15]. Indeed, under the assumption of exactness, employing Prop. 2.6 instead of Prop. 2.1 we replace (ii) by S > 0 such that d Gn (g, h) S ξ g, ξ h = 0. Using (4) and the surrounding discussion conclude that this property is shared by ξ. 4. Extensions Let 1 H Γ G 1 be an extension of countable discrete groups. We study uniform embeddability of Γ under various hypotheses on H and G. Our primary result, of which our other results are consequences, is the following theorem. Theorem 4.1. Let Γ be an extension of H by G as above. If H is uniformly embeddable and G is exact then Γ is uniformly embeddable. As corollaries we mention the following two results about semi-direct products. Corollary 4.2. Let G and H be countable discrete groups. Let α : G Aut(H) be an action of G on H. If both H and G are uniformly embeddable and α(g) Aut(H) is exact then the semi-direct product Γ = H G is uniformly embeddable.

10 10 MARIUS DADARLAT AND ERIK GUENTNER Proof. The semi-direct product Γ is the set of pairs (s, x) H G, with product (s, x)(t, y) = (sα x (t), xy). The assignment (s, x) ((s, α x ), x) : Γ (H α(g)) G. defines an injective homomorphism. By the theorem the semi-direct product H α(g) is uniformly embeddable, as is (H α(g)) G. In particular, Γ is a subgroup of the uniformly embeddable group and is therefore uniformly embeddable. Corollary 4.3. Let Γ = Z n G be a semi-direct product. If G is uniformly embeddable then Γ is uniformly embeddable. Proof. Apply the previous corollary using the fact that GL n (Z) is exact [15]. Remark. Central extensions are more difficult to analyze than semi-direct products. Our theorem applies to central extensions in which the quotient is exact. Gersten has shown that if a central extension of Z is described by a bounded cocycle then Γ is quasi-isometric to the product Z G [11]. Consequently, if G is uniformly embeddable so is Γ. Beyond these two results little is known. As remarked earlier, the property of uniform embeddability of a countable discrete group is independent of the proper length function. Consequently, we are free to choose these for our groups G, H and Γ in a convenient manner, which we do as follows. Let l Γ be a proper length function on Γ. Define length functions on H and G according to l H (s) = l Γ (s), for all s H, (5) l G (x) = min{ l Γ (a) : a Γ and ȧ = x }, for all x G, (6) where we have introduced the notation a ȧ for the quotient map Γ G. It is easily verified that the minimum in the definition of l G is attained, and that l G is a proper length function on G; that l H is a proper length function is immediate. Denote the associated left invariant metrics by d Γ, d G and d H, respectively. Observe that the inclusion H Γ is an isometry, and the quotient map Γ G is contractive. Finally, choose a set theoretic section σ of the quotient map Γ G with the property that and define η : Γ G H by l Γ (σ(x)) = l G (x), for all x G (7) η(a, x) = σ(x) 1 aσ(ȧ 1 x), for all a Γ, x G. (8)

11 Lemma 4.4. Let a, b Γ, x G. We have UNIFORM EMBEDDABILITY 11 d Γ (a, b) d G (x, ȧ) + d G (x, ḃ) + d H(η(a, x), η(b, x)) (9) d H (η(a, x), η(b, x)) d G (x, ȧ) + d G (x, ḃ) + d Γ(a, b). (10) Proof. Let a, b and x G be as in the statement. From the definition (8) of η we obtain, η(a, x) 1 η(b, x) = σ(ȧ 1 x) 1 a 1 b σ(ḃ 1 x). The desired inequalities follow easily from this equality, together with (5), (7) and the subadditivity of length functions. Proof of Thm We prove that Γ satisfies the conditions of Prop Let ε > 0 and R > 0 be given. We show that there exists an f : Γ l 2 (G, H) such that f(a) = 1 for all a Γ, and that satisfies the convergence and support properties: (i) if d Γ (a, b) R then 1 f(a), f(b) < ε (ii) ˆε > 0 S > 0 such that if d Γ (a, b) S then f(a), f(b) < ˆε. Adapting an argument of Anantharaman-Delaroche [3] our strategy is to use a g satisfying the conditions of Prop. 2.6 to average an h satisfying those of Prop. 2.1 to produce f. Since G is exact there exists, according to Prop. 2.6, a g : G l 2 (G) and an S G > 0 such that g(x) = 1 for all x G and such that (i g ) 1 g(x), g(y) < ε/2, provided d G (x, y) R, (ii g ) supp g(x) B SG (x), for all x G. We shall without comment view g as a function on G G whenever convenient. Since H is uniformly embeddable there exists, according to Prop. 2.1, an h : H H such that h(s) = 1 for all s H and such that (i h ) 1 h(s), h(t) < ε/2, provided d H (s, t) 2S G + R, (ii h ) ˆε > 0 S H > 0 such that if d H (s, t) S H then h(s), h(t) < ˆε. Having chosen g and h define f : Γ l 2 (G, H) by f(a)(x) = g(ȧ, x)h(η(a, x)), for all a Γ, x G. Note that f(a) l 2 (G, H). It is elementary to verify that f(a) = 1, for all a Γ. We verify the remaining properties.

12 12 MARIUS DADARLAT AND ERIK GUENTNER For the convergence property let a, b Γ with d Γ (a, b) R. Consider { 1 h(η(a, x), h(η(b, x)) } g(ȧ, 1 f(a), f(b) x)g(ḃ, x) + x G (11) 1 g(ȧ), g(ḃ). Since the map Γ G is contractive we have d G (ȧ, ḃ) R so that by (i g) the second term in (11) is bounded by ε/2. Observe that, according to (ii g ) the sum in the first term is over x B SG (ȧ) B SG (ḃ). Recalling that g(ȧ) = g(ḃ) = 1 we therefore bound the first term by sup{ 1 h(η(a, x)), h(η(b, x)) : x B SG (ȧ) B SG (ḃ)}. From (10) we see that for x B SG (ȧ) B (ḃ) we have d SG H(η(a, x), η(b, x)) 2S G + R so that by (i h ) this supremum, and consequently the first term in (11), is bounded by ε/2. For the support property let ˆε > 0 be given and obtain S H as in (ii h ). Let a, b Γ be such that d Γ (a, b) 2S G + S H. Then, f(a), f(b) = g(ȧ, x)g(ḃ, x) h(η(a, x), h(η(b, x)) x G x g(ȧ, x)g(ḃ, x) h(η(a, x), h(η(b, x)) sup{ h(η(a, x)), h(η(b, x)) : x B SG (ȧ) B SG (ḃ) }, where, again we use the fact that g(ȧ) = g(ḃ) = 1 to obtain the second inequality and note that by (ii g ) the sums are in fact over x B SG (ȧ) B SG (ḃ). From (9) we see that for such x we have d H (η(a, x), η(b, x)) d Γ (a, b) d G (x, ȧ) d G (x, ḃ) S H so that by (ii h ) the supremum is indeed bounded by ˆε. 5. Free Products The main result of this section is the following theorem. Theorem 5.1. Let A and B be countable discrete groups and let C be a common subgroup. If both A and B are uniformly embeddable then the amalgamated free product A C B is uniformly embeddable. Our strategy for proving the theorem is to construct a locally finite metric space X which is uniformly embeddable, and on which Γ acts freely by isometries. The construction of X is based on the notion of a tree of metric spaces, which we now recall.

13 UNIFORM EMBEDDABILITY 13 A tree T consists of two sets, a set V of vertices and a set E of edges, together with two endpoint maps E V associating to each edge its endpoints. Every two vertices are connected by a unique geodesic edge path, that is, a path without backtracking. A tree of spaces X (with base the tree T ) consists of a family of metric spaces { X v, X e } indexed by the vertices v V and edges e E of T together with maps σ e,v : X e X v whenever v is an endpoint of e. The σ e,v are the structural maps of X. We will assume, although this is not strictly necessary, that the metrics on the vertex and edge spaces are integer-valued. The total space X of the tree of spaces X is the metric space defined as follows. The underlying set of X is the disjoint union of the vertex spaces X v ; the metric on X is the metric envelope d of the partial metric d (see the appendix) defined by d(x, y), v V such that x, y X v d(x, y) = 1, e E and z X e such that x = σ e,v (z), y = σ e,w (z), for all (x, y) in the domain D = { (x, y) : x, y X v, v V } { (σ e,v (z), σ e,w (z)) : z X e, e E }. Observe that D is ample (see the appendix); this follows from our assumption that the underlying tree is connected, and that d is defined for all pairs (x, y) where x and y are in the same vertex space. We call (x, y) an adjacency if there exists an edge e, with endpoints v and w, and an element z X e, such that σ e,v (z) = x and σ e,w (z) = y. Using this terminology, the partial metric can be described as being the given metric on each vertex space and 1 on adjacencies. Example 5.2. Consider the case in which the vertex spaces are metric graphs (or rather the set of vertices of a graph, equipped with the graph metric) and the edges spaces are singletons. The total space X is itself a metric graph. Indeed, it is the disjoint union of the graphs, together with additional edges coming from the underlying tree. Precisely, the edges of X are first, the edges in the individual X v and second, the edges e of the underlying tree; the endpoints of such an edge e are the images of the maps X e X v to the endpoints v of e in T. Remark. The generality of the definition above is mandated by the fact that when considering amalgamated free products, we will encounter vertex spaces that are not metric graphs.

14 14 MARIUS DADARLAT AND ERIK GUENTNER Theorem 5.3. Let X be a tree of metric spaces in which the structural maps are isometries. If the vertex spaces X v are uniformly embeddable in a Hilbert space (with common distortion bound) then the total space X is uniformly embeddable in a Hilbert space. Remark. Quite explicitly, the hypothesis is that there exist maps ρ ± as in (1) such that for every vertex v V there exists F v : X v H satisfying ρ (d Xv (x, x )) F v (x) F v (x ) ρ + (d Xv (x, x )), for all x, x X v. In other words, the same distortion bounds may be used for every vertex space. Remark. If there exist only finitely many distinct isometry types of vertex spaces (as will be the case for the amalgamated free product) this simply means that every vertex space is uniformly embeddable. Remark. When we prove the theorem we will use the fact that the existence of a common distortion bound on the uniform embeddings of the vertex spaces implies that the estimates in the fundamental criterion for embeddability are uniform. This follows from the proof of Prop. 2.1, specifically from the estimate (2). Assuming Thm. 5.3 we prepare for the proof Thm. 5.1 by associating a tree of metric spaces X Γ to the amalgamated free product Γ = A C B. The tree T is the Bass-Serre tree of Γ [17, 4]. Precisely, the vertex and edge sets of T are given by V = Γ/A Γ/B E = Γ/C, respectively; the endpoint maps are the quotient maps Γ/C Γ/A and Γ/C Γ/B. In other words, the endpoints of an edge (a C-coset) are the vertices (one A and one B-coset) that contain it. We associate a metric space to each vertex and edge of T as follows. Equip Γ with an integer-valued proper length function and associated metric. Let v V be a vertex and assume that v Γ/A. In particular, v is an A-coset in Γ; denote by X v this coset itself, metrized as a subspace of Γ. Proceed similarly for vertices v Γ/B. Let e E be an edge. In particular, e is a C-coset in Γ; denote by X e this coset itself, again metrized as a subspace of Γ. The structural maps are defined as follows. Let the vertex v be an endpoint of the edge e. Inclusion of cosets (subsets of Γ) provides the structural map σ e,v : X e X v.

15 UNIFORM EMBEDDABILITY 15 Remark. The metric space X v is not (isometric to) the graph metric space of the Cayley graph of A or B, or even of the restriction of the Cayley graph of Γ to the subset A or B. Similar remarks apply the edge space X e. Indeed, it is important that we metrize X v and X e as subspaces of Γ and not via their identification with A or B and C using the given metrics on these groups. Let X Γ be the total space of X Γ. The group Γ acts on X Γ by left multiplication. Precisely, if x X v and a Γ we define a x X a v, using ordinary multiplication in Γ, by virtue of the fact that x v Γ. This action preserves adjacencies and, since in addition the metric on Γ is left invariant, it preserves the partial metric. According to Prop. 7.2 of the appendix Γ acts by isometries on the total space X Γ. Proposition 5.4. Let A and B be countable discrete groups and let C be a common subgroup. Let Γ = A C B be the amalgamated free product. Let X Γ be the tree of metric spaces associated to Γ and let X Γ be its total space. We have: (i) the structural maps of X Γ are isometries, (ii) every vertex space of X Γ is isometric to one of A or B (which are metrized with the subspace metric via the inclusions A, B Γ). (iii) The action of Γ by isometries on X Γ is free. (iv) X Γ is locally finite. Proof. All of the assertions but (iv) follow from the previous discussion. The condition (iv), while apparent, it will be derived formally in Prop Proof of Thm According to the previous proposition the tree of metric spaces X Γ associated to the amalgamated free product Γ = A C B satisfies the hypothesis of Thm. 5.3, according to which the total space X Γ is uniformly embeddable. Again according to the previous proposition Γ acts freely by isometries on X Γ. Hence, by Cor. 2.5, Γ is uniformly embeddable. We have reduced our main theorem concerning amalgamated free products, Thm. 5.1, to a theorem concerning trees of metric spaces, Thm We now turn attention to the proof of that theorem. We suitably adapt the method employed by Tu in his study of Property A for discrete metric spaces [19]. For a tree of metric spaces X the inclusions X v X of vertex spaces into the total space are, as a general rule, not isometric; the presence of shortcuts in neighboring X w may cause the distance in X between two vertices x, y X v to be considerably smaller than the

16 16 MARIUS DADARLAT AND ERIK GUENTNER distance between them in X v itself. Nevertheless, for our X Γ these inclusions are isometries, a fact that may be traced back to the manner in which the vertex and edge spaces of X Γ are metrized as subspaces of Γ which circumvents any distortion that may have otherwise been introduced by the amalgamating subgroup. The following proposition has no analog in Tu s work [19]; nevertheless, we require it in order to complete our arguments. Proposition 5.5. Let X = { X v, X e } be a tree of metric spaces in which the structural maps X e X v are isometries. Let X be the total space of X. Then (i) the inclusions X v X are isometries, and (ii) if (x, y) is an adjacency then d(x, y) = 1. Further, for all x X v, y X w d(x, y) = d T (v, w) + inf{ d(x 0, x 1 ) + d(x 2, x 3 ) + + d(x p 1, x p ) }, (12) where d T is the distance on the tree T and the infimum is taken over all sequences x 0,..., x p where p = 2d T (v, w) + 1, and (i) x = x 0, y = x p, (ii) (x 2k 1, x 2k ) is an adjacency for k = 1,..., d T (v, w), (iii) x 2k, x 2k+1 X vk, for k = 0,..., d T (v, w), and v = v 0,..., v dt (v,w) = w are the vertices along the unique geodesic path in T from v to w. Remark. Formula (12) is Tu s formula [19, Section 9] with slightly different notation. Proof. Let v, w V and let x X v and y X w. A reduced path from x to y is a sequence of elements x = x 0, x 1,..., x 2n, x 2n+1 of X for which there exist vertices v 0,..., v n V such that (i) v j v j+1, for j = 1,..., n, and (ii) x 2k, x 2k+1 X vk, for k = 0, 1,..., n. Observe that by appropriately inserting and deleting x s we may alter a path, without increasing its length, so as to obtain a reduced path (use the triangle inequality for the metrics on the individual X v ). Consequently, when computing distances in X it suffices to consider reduced paths. Let x = x 0,..., x 2n+1 = y be a reduced path from x to y. According to the definitions of a reduced path and of the domain D of the partial metric we obtain sequences of (i) vertices v = v 0,..., v n = w in V, (ii) edges e 1,..., e n in E, and

17 satisfying UNIFORM EMBEDDABILITY 17 (iii) elements z 1,..., z n of the edge spaces X e1,..., X en, (i) x 2k, x 2k+1 X vk, for k = 0,..., n, (ii) the endpoints of e k are v k 1 and v k, for k = 1,..., n, and (iii) σ ek,v k 1 (z k ) = x 2k 1 and σ ek,v k (z k ) = x 2k, for k = 1,..., n (The sequences are uniquely determined by these conditions.) The given reduced path x = x 0,..., x p = y in X lies over the edge path e 1,..., e n in T. We show that in the definition of d it suffices to consider reduced paths lying over the unique geodesic edge path in T from v to w; the assertions of the proposition follow easily from this fact. Let x = x 0,..., x 2n+1 = y lie over the non-geodesic path e 1,..., e n in T. We show that by successive elimination of certain x i we obtain a shorter path lying over the geodesic path in T. Indeed, there exists i 1,..., n 1 such that e i and e i+1 have the same endpoints, that is, such that v i 1 = v i+1. We have } } x 2i 2 x 2i X vi 1, X vi, x } 2i+2 X vi+1 = X vi 1. x 2i 1 x 2i+1 x 2i+3 We eliminate x 2i 1, x 2i, x 2i+1, x 2i+2 from the given path and claim that the resulting path (i) is shorter than x 0,..., x 2n+1, and (ii) lies over an edge path with fewer backtracks than e 1,..., e n. Of these (ii) is obvious; the new path lies over the edge path e 1,..., e i 1, e i+2,..., e n obtained by eliminating e i and e i+1. Further, (i) follows from the fact that the structural maps are isometries. In particular we have from which follows d(x 2i 1, x 2i+2 ) = d(z i, z i+1 ) = d(x 2i, x 2i+1 ), d(x 2i 2, x 2i+3 ) d(x 2i 2, x 2i 1 ) + d(x 2i 1, x 2i+2 ) + d(x 2i+2, x 2i+3 ) = d(x 2i 2, x 2i 1 ) + d(x 2i, x 2i+1 ) + d(x 2i+2, x 2i+3 ). Proposition 5.6. Let X be the total space of a be a tree of metric spaces X in which the structural maps are isometries. Assume that each vertex space X v is locally finite and there is a uniform bound for the number of adjacencies of each point in X. Then X is locally finite. Proof. In view of formula (12) a finite radius ball in X can intersect only finitely many vertex spaces.

18 18 MARIUS DADARLAT AND ERIK GUENTNER 6. Proof of Theorem 5.3 We may enlarge the tree of metric spaces (if necessary) such that the underlying tree T will contain an infinite geodesic ω starting at some basepoint. For every v V let α(v) V be such that the edge [v, α(v)] points towards ω. For each v V let Y v = σ e,v (X e ) X v and f v = σ e,α(v) σe,v 1 : Y v X α(v) where e = [v, α(v)]. It follows immediately from Prop. 5.5 that each f v is isometric. Using the same proposition, and the notation just introduced, we rewrite the distance formula (12) as follows. Let x 0 X v and x 0 X v. There exist a unique pair of nonnegative integers k, l such that α k (v) = α l (v ) and d T (v, v ) = k + l, where again d T denotes the distance in the tree T. By symmetry we may assume that k l. If k 1 and l 1 then d(x 0, y 0 ) + d(f α i (v)(y i ), y i+1 ) i=0 d(x 0, x 0) = k + l + inf + d(f α (v)(y k 1 k 1 ), f α l 1 (v )(y l 1)) k 2 + l 2 j=0 d(f α j (v )(y j), y j+1) + d(y 0, x 0), (13) subject to the constraints that y i Y α i (v) and y j Y α j (v ). If k 1 and l = 0 then d(x 0, x 0) = k + inf { d(x 0, y 0 ) + k 2 i=0 d(f α i (v)(y i ), y i+1 ) + d(f α k 1 (v)(y k 1 ), x 0)) }, (14) subject to similar constraints. An n-chain is a sequence x = (x 0, x 1,..., x n 1 ) such that for each 0 k n 2 there exists x k Y α k (v) satisfying d(x k, x k ) < d(x k, Y α k (v)) + 1 and f α k (v)( x k ) = x k+1. If x 0 X v the n-chain starts in X v. Note that for any n 1, v V and x 0 X v there exists an n-chain whose initial element is x 0. Lemma 6.1. Let x 0 X v and x 0 X v, v v with d(x 0, x 0) < R and let k and l be as in (13) (14) with k l. If l = 0, there exists a chain (x 0, x 1,..., x k ) such that {( max sup 0 i k 1 d(x i, x i+1 ) ) }, d(x k, x 0) < 2 R R.

19 UNIFORM EMBEDDABILITY 19 If l 1, there exist chains (x 0, x 1,..., x k ), (x 0, x 1,..., x l ) and z, z X α k (v) = X α l (v ) such that {( ) ( ) } max sup d(x i, x i+1 ), sup d(x j, x j+1), d(x k, z), d(x l, z ), d(z, z ) < 2 R R. 0 i k 1 0 j l 1 Remark. In the proof of the lemma, and at a number of subsequent points, we require the fact that if x X v and y Y v then d(x, f v (y)) = d(x, y) + 1. This follows from Thm Indeed, since (y, f v (y)) is an adjacency, d(x, f v (y)) d(x, y)+1. For the reverse inequality, let ε > 0 and obtain x Y v such that d(x, f v (y))+ε 1+d(x, x )+ d(f v (x ), f v (y)). Since f v is an isometry, d(x, x ) + d(f v (x ), f v (y)) d(x, y) and we are done. Proof. Case l = 0: We have k 1 since v v and k l. Since d(x 0, x 0) < R, by (14) there is y 0 Y v such that k + d(x 0, y 0 ) < R. The sequence x 1,..., x k is constructed inductively. Let x 0 Y v be such that d(x 0, x 0 ) d(x 0, y 0 ) and d(x 0, x 0 ) < d(x 0, Y v ) + 1 and define x 1 = f v ( x 0 ). Then d(x 0, x 1 ) = d(x 0, x 0 ) + 1 d(x 0, y 0 ) + 1 < R. Thus d(x 0, x 1 ) < R and d(x 1, x 0) d(x 0, x 0) + d(x 0, x 1 ) < 2R. Repeating the same argument for the pair of points x 1, x 0 (if k 2) we find x 2 X α 2 (v) with (x 0, x 1, x 2 ) being a chain, d(x 1, x 2 ) < 2R and d(x 2, x 0) < 2 2 R. Continuing in the same way we obtain a chain (x 0, x 1,..., x k ) such that d(x i 1, x i ) < 2 i 1 R and d(x i, x 0) < 2 i R, 1 i k. Since k R this completes the proof of (a). Case l 1: Let y 0,..., y k 1 with y i Y α i (v) and y 0,..., y l 1 with y j Y α j (v ) be sequences such that the expression whose infimum is taken in (13) is less than R k l. Let z = f α l 1 (v )(y l 1 ) and z = f α (v)(y k 1 k 1 ). Then we have d(z, z ) < R, d(x 0, z) < R and d(x 0, z ) < R. The proof is completed by applying part (a) to the pairs x 0, z and x 0, z. Given an n-chain x = (x 0, x 1,..., x n 1 ) starting in X v define, for 0 k n 1, δ k = d(x k, Y α k (v)), and θ k = δ 0 δ k, where we introduce the notation a b = max{a, b}. For future notational convenience define θ 1 = 0. Note that if x = (x 0, x 1,..., x n 1) is another n-chain starting in X v then d(x k 1, x k ) δ k 1 + 2, (15) δ k δ k = d(x k, Y α k (v)) d(x k, Y α k (v)) d(x k, x k), 0 k n 1. (16)

20 20 MARIUS DADARLAT AND ERIK GUENTNER Lemma 6.2. Let x = (x 0, x 1,..., x n 1 ) and x = (x 0, x 1,..., x n 1) be n-chains starting in X v. Then d(x k, x k) d(x 0, x 0) 2k(θ k 1 θ k 1) + 2k, 0 k n 1. (17) Proof. Since the statement is obvious in the case k = 0 we assume k 1. Let x k and x k be as in the definition of n-chains. Observe that d( x k, x k) d( x k, x k ) + d( x k, x k) + d(x k, x k) d(x k, x k) + 2(θ k θ k) + 2, d( x k, x k) d(x k, x k) d( x k, x k ) d( x k, x k) d(x k, x k) 2(θ k θ k) 2. Since f α k (v) is an isometry we have d(x k+1, x k+1 ) = d( x k, x k ) and the lemma follows from these inequalities by induction. Given an n-chain x = (x 0, x 1,..., x n 1 ) and N > 0 we define, for 0 k n 1, ( a k = 1 ln(1 + θ ) k), (18) N + c k = a 0 a k 1 (1 + (n k 1)(1 a k )), (19) where we introduce the notation a + = max{a, 0}. Note that c 0 = 1 + (n 1)(1 a 0 ). One checks immediately that a 0 a 1 a n 1, c c 2 n 1 = n, 0 c n 1 1, (20) δ k e N = θ k e N = a k = 0 = c k+1 = = c n 1 = 0. (21) The coefficients c k were introduced by Tu [19] (actually we work with the square root of Tu s coefficients). Since the maps f v are isometries, it is possible in our case to define a k explicitly as in (18). Both a k and c k should be regarded as functions a x k and cx k of n-chains x = (x 0, x 1,..., x n 1 ). We will often write a k and c k instead of ax k and cx k. Lemma 6.3. Let x = (x 0, x 1,..., x n 1 ) and x = (x 0, x 1,..., x n 1) be n-chains starting in X v. If ω = max 0 i n 1 a i a i then (i) c 2 k c k 2 ( ) n ω, (ii) c k c k n+1 2 ω, and (iii) n 1 k=0 1 n c k c k 2 (n+1)2 2 ω. Proof. This is an exercise. For (i) (ii) use the inequality a b 2 a b. The following continuity property of the coefficients a k is a minor variation of a formula in Tu s paper [19, formula 9.2].

21 UNIFORM EMBEDDABILITY 21 Lemma 6.4. Let x = (x 0, x 1,..., x n 1 ) and x = (x 0, x 1,..., x n 1) be n-chains starting in X v and assume d(x 0, x 0) n. Then max a k a k 7n2 0 k n 1 N Proof. Denote λ = ln 7n. We are going to show that (22) a 0 a 0 n N a k a k a k 1 a k 1 + λ N. (23) From these we immediately obtain a k a n(1 + λ) k 7n2, concluding the proof. Of N N these inequalities, the first is straightforward. Indeed, using (16) and the fact that the map t (1 1 ln(1 + t)) N + is 1 -Lipschitz we have N a 0 a 0 1 N δ 0 δ 0 1 N d(x 0, x 0) n N. To prove the second inequality denote f(t) = 4nt + 6n. According to (16) and Lemma 6.2, we have δ k δ k d(x k, x k) 1 2 f(θ k 1 θ k 1). (24) Denote ψ(t) = ln(1 + t), so that we have for all t 0 ψ(t) ψ(f(t)) ψ(t) + λ. (25) Case θ k θ k f(θ k 1 θ k 1 ) : By symmetry we may assume that θ k 1 θ k 1. Thus θ k 1 θ k f(θ k 1) and θ k 1 θ k 1 θ k f(θ k 1 ). Using (25) we obtain ψ(θ k 1) ψ(θ k) ψ(f(θ k 1 )) ψ(θ k 1 ) + λ, ψ(θ k 1) ψ(θ k ) ψ(f(θ k 1 )) ψ(θ k 1 ) + λ. Now, (23) follows immediately from these inequalities, together with the fact that for real numbers a s, t b + λ we have (1 s/n) + (1 t/n) + (1 b/n) + (1 a/n) + + λ. Case θ k θ k f(θ k 1 θ k 1 ) : By symmetry we may assume that θ k θ k. Then θ k f(θ k 1 ) θ k 1 hence θ k = δ k. Therefore using (24) θ k θ k θ k + θ k δ k δ k + θ k 1 2 f(θ k 1 θ k 1) + θ k 1 2 θ k + θ k,

22 22 MARIUS DADARLAT AND ERIK GUENTNER hence θ k 2θ k f(θ k ). From θ k θ k f(θ k ) and (25) we obtain ψ(θ k) ψ(θ k ) ψ(f(θ k)) ψ(θ k) + λ. From this inequality, and the fact that the map t (1 t/n) + is 1 -Lipschitz we obtain N a k a k λ, which implies (23). N Definition. Given R > 0 and ε > 0 choose and fix n N such that ( ) 1 ln n ε < 4n 3(R + 3), n > 2R R, (26) and N N such that 6n(n + 1) N < ε 3(R + 3). (27) Having done so apply the fundamental criterion for uniform embeddability, Prop. 2.1, to choose and fix a family (ξ x ) x X of unit vectors in a Hilbert space H X = v V H v with ξ x H v if x X v and such that sup{ ξ y ξ y : d(y, y ) n(2e N + 3), y, y X v, v V } < ε 3(R + 3), (28) lim sup{ ξ y, ξ y : d(y, y ) S, y, y X v, v V } = 0. (29) S (See the remarks after Thm. 5.3 for comments on why this is possible.) Finally, for every n-chain x = (x 0, x 1,..., x n 1 ) in X define the unit vector η x H X by η x = 1 n 1 n k=0 c k ξ xk, (30) where, of course, the c k s are defined according to (19) and depend on the chain x. Lemma 6.5. Let x = (x 0, x 1,..., x n 1 ) and x = (x 0, x 1,..., x n 1) be n-chains starting in X v. If d(x 0, x 0) n, then η x η x 2ε. 3(R+3) Proof. Let I = {k : c k 0, c k 0} and J = {k : c k = 0, c k 0}. If 1 k I, then a k 1 = ( 1 1 ln(1 + θ N k 1) ) 0 hence θ + k 1 e N. Similarly, θ k 1 en. Hence from (17) we have d(x k, x k) 2ne N + 2n + d(x 0, x 0) n(2e N + 3). (31)

23 UNIFORM EMBEDDABILITY 23 Consider η x η x = 1 n 1 c k ξ xk 1 n 1 c n k=0 n k ξ x k k=0 1 n 1 (c k c 1 n k) ξ xk + (c k=0 n k c k ) (ξ xk ξ x k ) k J 1 + c n k (ξ xk ξ x k ). k I Observe that the ξ xk s and ξ x k s are in orthogonal components of H X. We bound the third term on the right using (20), (31) and (28) as follows: ( ) 1/2 ( 1 (c n k) 2 ξ xk ξ x k 2 1 n 1 k I n k=0 ) 1/2 ( (c k) 2 sup ξ y ξ y ε 3(R + 3), sup ξ xk ξ x k k I where = {(y, y ) : d(y, y ) n(2e N + 3), y, y X v, v V }. We bound the sum of the first two terms on the right, by ) 3 ( n 1 ) n c k c k 2 k=0 3(n + 1) 2 2 ( 7n 2 max a i a i 1 3(n + 1) 2 0 i n 1 N ) 1 2 ε 3(R + 3), where the inequalities are from Lemma 6.3, (22) and (27), respectively. Combining these observations we obtain the result. Lemma 6.6. Let x = (x 1, x 2,..., x n ) and x = (x 0, x 1,..., x n 1) be n-chains with x 0 X v and x 1 X α(v). If (x 0, x 1 ) is a 2-chain and d(x 0, x 1 ) n, then η x η x ε. R+3 Proof. Let x 0 = fv 1 (x 1 ) and set x = ( x 0, x 1,..., x n 1 ). According to the definition (30) of η x we have ( η x = η (x 1,x 2,...,x n) = 1 n 1 ) d k ξ xk + d n ξ xn, n k=1 ( η x = η ( x 0,x 1,...,x n 1 ) = 1 n 1 d 0 ξ x0 + n k=1 d k ξ xk ),

24 24 MARIUS DADARLAT AND ERIK GUENTNER where d k = c k 1 (x 1,..., x k ) = ā 1 ā k 1 (1 + (n k)(1 ā k )), d k = c k ( x 0, x 1,..., x k ) = ā 0 ā 1 ā k 1 (1 + (n k 1)(1 ā k )), { ( ) } with ā 0 = 1, d0 = 1 and ā k = a k ( x 0, x 1,..., x k ) = min 1 ln(1+δ i) : 1 i k. For N + 1 k n 1 we have either 1 ā k = 0, hence d k d k = 0, or ā1 d k d ā k 1 (1 ā k ) k = 1 + (n k)(1 āk ) (n k 1)(1 ā k ) (1 ā k ) 2 (n k)(1 ā k ) 1 2 n k. Applying the above expressions for η x and η x with this inequality we estimate η x η x = 1 ( d n 1 0 ξ x0 + ( d k d k )ξ xk d n ξ xn n k=1 ( ) 1 2 n + 1 n 1 (d k n d 2 k ) 2 k=1 ( 2 n + 1 ( 1 4n n n ( ) 1 ln n ε 4n 3(R + 3), )) 1 2 where the final inequality comes from the choice (26) of n. Apply Lemma 6.5 to the chains x and x, noting that d(x 0, x 0 ) = d(x 0, x 1 ) 1 n, and use the previous inequality to conclude η x η x η x η x + η x η x 2ε 3(R + 3) + ε 3(R + 3) = ε R + 3. Lemma 6.7. Let x = (x 0, x 1,..., x n 1 ) and x = (x 0, x 1,..., x n 1) be n-chains in X. Then η x, η x sup { ξ y, ξ y : y, y X v, v V, d(y, y ) d(x 0, x 0) 6n(e N + 1)}. Proof. Let k and l be as in (13). Considering symmetry we assume that k l. Also, if k n, then η x, η x = 0 so we assume that k < n. With these assumptions η x, η x = 1 n n k 1 i=0 c k+i c l+i ξ xk+i, ξ x l+i. (32)

25 UNIFORM EMBEDDABILITY 25 Making use of (21), conclude that if θ k 1 θ l 1 en then all the products c k+i c l+i in (32) are zero and we are done. Thus we assume that θ k 1 θ l 1 en. Let m be the largest number 0 m n k 1 with the property that θ k+m 1 θ l+m 1 en. By (17) d(x k+i, x l+i) d(x k, x l) 2n(e N + 1) for all 0 i m. On the other hand using (15) d(x k, x l) d(x 0, x 0) k 1 i=0 d(x i, x i+1 ) l 1 j=0 k 1 l 1 d(x 0, x 0) (δ i + 2) (δ j + 2) i=0 j=0 d(x 0, x 0) (k + l)(2(θ k 1 θ l 1) + 2) d(x 0, x 0) 4n(e N + 1). Combining these two inequalities we obtain d(x j, x j+1) d(x k+i, x l+i) d(x 0, x 0) 6n(e N + 1), for all 0 i m. Finally, arguing again on the basis of (21) as above, conclude that the terms in (32) for i > m are zero. Thus, applying (20) and the previous inequality, we see that ( ) η x 1 m, η x c k+i c l+i sup ξ y, ξ y sup ξ y, ξ y, n Ω Ω i=0 where Ω = {(y, y ) : y, y X v, v V, d(y, y ) d(x 0, x 0) 6n(e N + 1)}. Conclusion of proof of Theorem 5.3. Given R > 0 and ε > 0 let n, N, and (ξ x ) x X be constructed as in the Definition. For each x 0 X choose and fix an n-chain x = (x 0, x 1,..., x n 1 ) and consider the corresponding vector η x. We are going to show that the map x 0 η (x 0,x 1,...,x n 1 ) satisfies the convergence and support conditions of Prop The support condition follows immediately from Lemma 6.7 in conjunction with (29). For the convergence condition we show that for any x 0, x 0 X with d(x 0, x 0) < R and any two n-chains x and x starting at x 0 and x 0, respectively, we have η x η x ε. Let k and l be as in(13) (14). We may assume that k l. Furthermore we treat only the case l 1, the case l = 0 being similar. Let x(i), x (j), z and z be n-chains whose initial elements are the points x i, x j, z, z given by Lemma 6.1, 0 i k, 0 j l. Applying Lemmas 6.5 and

26 26 MARIUS DADARLAT AND ERIK GUENTNER 6.6 repeatedly (note that since n > 2 R R this is possible by Lemma 6.1), with the convention that all empty sums are zero, we have k 1 η x(i) η x(i+1) + η x(k) η z + η z η z η x η x i=0 l 1 + η z η x (l) + η x (j) η x (j+1) (k + l + 3)ε R + 3 ε. Remark. In the proof of Thm. 5.3 we showed that for every R > 0 and every ε > 0 there is a family (η x ) x0 X of unit vectors in H X such that j=0 sup{ η x η x : d(x 0, x 0) < R} ε, η x, η x sup { ξ y, ξ y y, y X v, v V, d(y, y ) d(x 0, x 0) 6n(e N + 1)}. Let us argue that, in addition to Thm. 5.3, these estimates prove that A C B is exact whenever A and B are exact [8, 9, 19]. Indeed, if we base the construction (30) of η x on a family (ξ x ) x X satisfying the support condition (ii) of Prop. 2.6 uniformly with respect to v V, then it follows from the second inequality that η x, η x = 0 whenever d(x 0, x 0) S +6n(e N +1). In particular, X satisfies the convergence and support conditions of Prop. 2.6 and A C B is exact by Cor Appendix We collect several elementary results on the construction of metrics required for our treatment of amalgamated free products. Let X be a set. A partial metric on X is an integervalued function d : D R + defined on a domain D X X satisfying, for all x, y X (i) d(x, y) = d(y, x) (ii) d(x, y) = 0 if and only if x = y In these statements it is assumed that all relevant pairs belong to the domain D of d and that the domain D is symmetric and contains the diagonal. We now associate a metric to a partial metric. The construction is analogous to that of path metrics; intuitively the distance between two points is the length of the shortest path between them. A path from x to y X is a sequence x = x 0, x 1,..., x n = y such that for every 1 j n we have (x i 1, x i ) D. The length of a path is n d(x i=1 i 1, x i ). The domain D is ample if for every x, y D there exists a path from x to y.